what is the meaning of "implicit declaration of function `malloc'"?

[nick]

the following is my programming code and compile message
why the warning message arise, have i done somethings wrong?


#include<stdio.h>
typedef struct card{

int abc;
}card;

int main(){


card *d;
d = (card*) malloc(sizeof(card));


return 0;

}

the compile message :

--------------------Configuration: test - Debug--------------------
Compiling source file(s)...
test.c
test.c: In function `main':
test.c:11: warning: implicit declaration of function `malloc'
Linking...

test.exe - 0 error(s), 1 warning(s)

thanks!

 

[pete]

the following is my programming code and compile message
why the warning message arise, have i done somethings wrong?

#include<stdio.h>

You neglected to write this here:

d = (card*) malloc(sizeof(card));

test.c:11: warning: implicit declaration of function `malloc'

The declaration for malloc, is in stdlib.h.
Without the declaration in scope,
C89 assumes malloc returns type int.

With the declaration in scope,
the compiler knows that malloc returns type pointer to void.
The (card*) cast that you used, suppressed the warning
that you would have gotten for assigning an int value
to a pointer type.

Add
#include <stdlib.h>
and lose the cast.

 

[Martin Ambuhl]

the following is my programming code and compile message
why the warning message arise, have i done somethings wrong?

The error message

test.c:11: warning: implicit declaration of function `malloc'

tells you: you have failed to have a declaration for the function
'malloc' in scope. This is easily fixed with
#include <stdlib.h>

#include<stdio.h>
typedef struct card{
int abc;
}card;

int main(){
card *d;
d = (card*) malloc(sizeof(card));

There is no excuse for the cast. If you put it in to stop complaints
from the compiler, you treated the symptom rather than the problem.
The normal form of this call is
d = malloc(sizeof *d);
And you should check the returned value.

> return 0;
> }

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